# Solve the equation ?

## Solve the equation ? ${\left(z - 1\right)}^{3} = 2 - 11 i$

Jan 17, 2018

${x}^{3} - i {y}^{3} + 3 i {x}^{2} y - 3 x {y}^{2} - 3 {x}^{2} + 3 {y}^{2} - 6 i x y + 3 x + 3 i y$$= 3 - 11 i$

#### Explanation:

given ${\left(z - 1\right)}^{3} = 2 - 11 i \to \left(1\right)$ expanding it we get
${z}^{3} - 1 - 3 {z}^{2} + 3 z = 2 - 11 i \to \left(2\right)$
let us assume $z = x + i y \to \left(3\right)$ put $\left(3\right)$ in $\left(2\right)$
${\left(x + i y\right)}^{3} - 3 {\left(x + i y\right)}^{2} + 3 \left(x + i y\right) = 3 - 11 i$
${x}^{3} - {y}^{3} i + 3 i {x}^{2} y - 3 x {y}^{2} - 3 \left({x}^{2} - {y}^{2} + 2 i x y\right) + 3 \left(x + i y\right)$ $= 3 - 11 i$
${x}^{3} - i {y}^{3} + 3 i {x}^{2} y - 3 x {y}^{2} - 3 {x}^{2} + 3 {y}^{2} - 6 i x y + 3 x + 3 i y$$= 3 - 11 i$